# Hyperbolic Geometry Lecture 2 by ild18893

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```									                   Hyperbolic Geometry Lecture 2
John Stogin
October 3, 2009

First, we begin with a couple comments from last week. There are two
matrices in SL(2, R) that correspond with the same isometry of H, and their
coordinates diﬀer by a sign. So we will work with P SL(2, R) = SL(2, R)/±12
as our set of isometries of H. Also, P SL(2, R) is a subset of a special class
of functions called conformal maps. These maps satisfy the Cauchy-Riemann
∂v               ∂v
equations given by ∂u = ∂y and ∂u = − ∂x for u + iv = T (x + iy). They also
∂x            ∂y
have the property that their derivative is never zero and they preserve angles.

1        Hyperbolic Geometry on the Unit Disk
1.1        Introducing U
Last week, we saw how the hyperbolic geometry can be represented as a half
2
+dy 2
plane, H = {z ∈ C|Im(z) > 0} with a metric ds2 = dx y2 . Today, we will
also consider another representation.
Deﬁnition (The Unit Disk) We call U = {z ∈ C||z| < 1} the unit disk.
There is a conformal map f : H → U given by f (z) = zi+1 . Note that this is a
z+i
bijection and maps the boundary R ∪ {∞} of H to the boundary {z||z| = 1} of
U. We also deﬁne a distance du on U by du (z0 , z1 ) = dh (f −1 z0 , f −1 z1 ) where
dh is the distance on H.
4|dz|2
Theorem 1.1 The distance du can be calculated with the metric ds2 =               (1−|z|2 )2

Proof We will show this by calculating the length of a curve γ : [a, b] → U.
−1
b | d f ◦γ(t)|dt
So lu (γ) = lh (f −1 ◦ γ) = a dt Im(γ)         . In order to do this, we must
−1
ﬁrst calculate f     and we do so by solving w = f (z) = zi+1 for z. So
z+i
wz + iw = zi + 1 ⇒ z(w − i) = −iw + 1 ⇒ z = f −1 (w) = −iw+1 . Now, we cal-
w−i
−iγ+1
culate the denominator of the integral expression. Im(f −1 γ) = Im               γ−i      =
(−iγ+1)(¯ +i)
γ           Im(−iγ γ +γ+¯ +i)
¯    γ           1−|γ|2
Im                       =       |γ−i|2          =   |γ−i|2 .   And the numerator becomes
(γ+i)(γ+i)
d      −1                   −1
dt f      ◦ γ(t) dt = |(f     ) ◦ γ(t)||γ (t)|dt and for the sake of simplicity, we will
now evaluate      d −1
dz f (z)    =   d
dz
−iz=1
z−i   = −i(z−i)−(−iz+1) = (z−i)2 . So the entire
(z−i)2
−2

b               2   b 2|dz|2
|γ−i|
integral expression now        becomes a 2|γ (t)|dt 1−|γ|2 = a 1−|z|2 where z = γ(t).
|γ−i|2

For an example, let us calculate verify the following corollary.

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Corollary 1.2 The distance between a point in U with radius r and the origin
r+1
is ln r−1 .

Proof We can choose our point to be a real positive number without any
r 2dx    r 1      1                     r
loss of generality. du (0, r) = 0 1−x2 = 0 1−x − 1+x dx = [− ln |1 − x|]0 +
r              1+r
[− ln |1 + x|]0 = ln             1−r    .

1.2     Two related lemmas
|z−w|2
Lemma 1.3 For z, w ∈ H, cosh(dh (z, w)) = 1 +                                2Im(z)Im(w) .

b
Proof We know dh (ai, bi) = ln a assuming without loss of generality that a < b.
2
+a2
Let’s verify the formula in this case. cosh(dh (ai, bi)) = b/a+a/b = b 2ab =
2
2            2                               2
|ai−bi|
2ab
2ab + b −2ab+a = 1 + 2Im(ai)Im(bi) . Now, we know that distance is invari-
2ab
ant under Mobius transformations, so if the right hand side is also invariant,
then we can prove the equation for any z and w by selecting a Mobius trans-
formation T mapping both of them to the imaginary axis. First let’s evaluate
2                                  2
|T z − T w|2 =            cz+d   −     cw+d       =         (cz+d)(cw+d)             =   |cz+d|2 |cw+d|2 .     At this
point, we only have to evaluate Im(T z) = Im                              az+b
cz+d
z +d)
= Im (az+b)(c¯+d) =
z
(cz+d)(c¯
¯        z
|cz+d|2
z
|cz+d|2
|cz+d|2
Im(z)
= |cz+d|2 . We now see that
we will get proper cancellation and that the right side of our equation is also
invariant.
2|z−w|2
Lemma 1.4 In U, lemma (1.3) becomes cosh(du (z, w)) = 1 +                                           (1−|z|2 )(1−|w|2 ) .

Proof The proof of this lemma is similar to the proof of lemma (1.3). It is
not diﬃcult to show that f −1 (z) = −iz+1 . From lemma (1.3), cosh(du (z, w)) =
z−i
|f −1 z−f −1 w|2
cosh(dh (f −1 z, f −1 w)) = 1 +               2Im(f −1 z)Im(f −1 w) .        We can evaluate the numera-
2                       2
−iz+1       −iw+1
tor of this expression: |f −1 z − f −1 w|2 =                          z−i    −    w−i           =       2w−2z
(z−i)(w−i)        =
4|z−w|2
At this point, we only have to evaluate Im(f −1 z) to ensure that
|z−i|2 |w−i|2 .
we will get proper cancellation in our fraction. Let’s do that. Im(f −1 z) =
−iz+1                              z
(−iz+1)(¯+i)                        ¯    z
Im(−iz z +z+¯+i)       1−|z|2
Im     z−i        = Im                                    =       |z−i|2         =   |z−i|2 ,   which is what we
(z−1)(z−i)
want.

2       Hyperbolic Trigonometry
2.1     Angle of Parallelism
Consider a triangle ∆ in H with a vertex at ∞ and another vertex and right
angle at i.
Deﬁnition We deﬁne Π(a) to be the third angle of ∆ with a side of length a
opposite the vertex at ∞.
We can now explore the following relations.

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Theorem 2.1 Given ∆ satisfying the conditions mentioned above, the following
three relations hold.
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(i) tan Π(a) =
sinh(a)
1
(ii) sin Π(a) =
cosh(a)
1
(iii) sec Π(a) =
tanh(a)

Proof From lemma (1.3), let the third vertex of ∆ be v so we can validate (ii).
Π(a)
|i−v|2           4 sin2 ( π − 2 )
4         sin Π(a)+(1−cos( π −Π(a)))
2
cosh(a) = 1 + 2Im(i)Im(v) = 1 +       2 sin Π(a)   =           sin Π(a)         =
1
sin Π(a) .
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To validate (iii), note sec Π(a) = cos Π(a) =                               1 − sin2 Π(a) =       1−         1
cosh2 (a)
=
√
cosh2 (a)−1   sinh(a)
cosh(a)    = cosh(a) = tanh(a).
Now, (i) is trivial.

2.2        The Sine and Cosine Rules
In our new hyperbolic trigonometry, there are some formulae that are similar to
those that we have seen in trigonometry. As usual in trigonometry, we concern
ourselves with the study of triangles with sides measuring a, b, and c, and having
opposite angles α, β, and γ respectively.
Theorem 2.2 (Cosine Rule I) cosh(c) = cosh(a) cosh(b)−sinh(a) sinh(b) cos(γ)

Proof For this proof, we will deﬁne a triangle ∆ in U with vertices vc = 0,
r = Re(va ) and Im(va ) = 0, and z = vb and opposite lengths c, a, and b respec-
tively. Note that there is no loss of generality here. From lemma (1.4), cosh(c) =
2|va −vb |2
(1−|va |2 )(1−|vb |2 ) + 1. We will now verify the right side matches this expression.
1+r   1−r
1−r + 1+r          2(1+r 2 )
Note using corollary (1.2), cosh(b) = cosh(du (0, r)) =                                  2         =     2(1−r 2 )    =
2
1+r 2                                                                                                  2
1−r 2 .     Now we can also ﬁnd sinh(b) =                             cosh2 (b) − 1 =             1+r
1−r 2       −1 =
√
(1+r 2 )2 −(1−r 2 )2
2r
1−r 2    = 1−r2 . And from the traditional cosine rule, |z − r|2 = |z|2 +
2
r −2|z|r cos(γ). Now, we can directly calculate the right side. cosh(a) cosh(b)−
2                   2    2       2
1+r 2 1+|z|
sinh(a) sinh(b) cos(γ) = 1−r2 1−|z|2 − (1−r2 )(1−|z|2 ) r +|z| −|z−r|
4r|z|
2r|z|
2        2        2      2         2
r 2 |z|2 −r 2 −|z|2 +1+2|z−r|2               2|z−r|2
= (1+r )(1+|z| )−2r −2|z| +2|z−r| =
(1−r 2 )(1−|z|2 )                                     (1−r 2 )(1−|z|2 )       = 1+     (1−r 2 )(1−|z|2 ) ,
which is what we wanted.
sinh(a)         sinh(b)
Theorem 2.3 (Sine Rule)                    sin(α)   =     sin(β)

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sinh(c)            sinh2 (c)                     sinh2 (c)
Proof           sin(γ)       =   1−cos2 (γ)   =                 2 from theorem (2.2).
cosh(a) cosh(b)−cosh(c)
1−(          )
sinh(a) sinh(b)
We will now show the right side is symmetric in a, b, and c. It can be rewrit-
sinh2 (a) sinh2 (b) sinh2 (c)
ten as sinh2 (a) sinh2 (b)−(cosh(a) cosh(b)−cosh(c))2 , which is symmetric provided the
denominator is symmetric. Using cosh2 (x) = 1 + sinh2 (x) and expanding, we

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see the denominator is equivalent to
sinh2 (a) sinh2 (b) − cosh2 (a) cosh2 (b) − cosh2 (c) + 2 cosh(a) cosh(b) cosh(c) =
sinh2 (a) sinh2 (b)−(1+sinh2 (a)) cosh2 (b)−cosh2 (c)+2 cosh(a) cosh(b) cosh(c) =
sinh2 (a)(sinh2 (b) − cosh2 (b)) − cosh2 (b) − cosh2 (c) + 2 cosh(a) cosh(b) cosh(c).
Since sinh2 (b) − cosh2 (b) = −1, we are done.
cos(α) cos(β)+cos(γ)
Theorem 2.4 (Cosine Rule II) cosh(c) =                         sin(α) sin(β)

Proof This proof is rather straightforward and tedious. It is therefore left as
an exercise.

Note that this third theorem has no analogue in trigonometry on the Euclidean
plane. Can you see why? It determines the length of a side from the three
angles, whereas in Euclidean trigonometry, there are many diﬀerent triangles
with congruent angles.

3    Hyperbolic Area
Since we have explored the notion of length, and we are working in a two
dimensional space, it seems only natural to explore the concept of area next.
dxdy
Deﬁnition (Area) For a region A ⊂ H, deﬁne the area to be µ(A) =                           A    y2    if
the integral exists.
This is a natural deﬁnition of area as we will see in the following theorem.
Theorem 3.1 The area µ(A) is invariant under transformations in P SL(2, R).
I.e. for T ∈ P SL(2, R), µ(T (A)) = µ(A).
az+b
Proof First deﬁne T (z) =             cz+d   where ad−bc = 1. Let w = u+iv = T (z). Then
∂u ∂v       ∂u ∂v                  ∂u 2         ∂v 2             ∂           2
dudv =       ∂x ∂y   −   ∂y ∂x       dxdy =     ∂x     +     ∂x     dxdy =    ∂x (u   + iv) dxdy =
2                   2
∂
∂x T z(x) dxdy = dT dxdy = |cz + d|−4 dxdy. The second equality holds by
dz
the Cauchy-Reimann equations. Then, recalling our proof of lemma (1.3) that
Im(z)
v = Im(T (z)) = |cz+d|2 , we calculate
dudv           |cz+d|−4 dxdy             dxdy
µ(T (A)) =     T (A)      v2    =    A Im(z)2 |cz+d|−4
=    A    y2    = µ(A)

4    Gauss-Bonnet Theorem
Theorem 4.1 (Gauss-Bonnet) Let ∆ be a triangle with angles α, β, and γ.
Then µ(∆) = π − α − β − γ.

Proof First, we will observe the case where a vertex of the triangle lies on
the boundary ∂H and from theorem (3.1) use a mobius transformation to map
this vertex to ∞. This yields a region bounded on the left and right by two
vertical lines and below by a subset of a circle. Without changing angles and
area, we may apply transformations of the form T (z) = z + b or T (z) = az so
that this circle is the unit circle. (Let the left side be adjacent to angle α and
have horizontal coordinate a, while the right side is adjacent to angle β and has
horizontal coordinate b.) Finally, we have a region over which it is not diﬃcult

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to integrate.
b        ∞                b
µ(A) = A dxdy =
y2      a
dx   √      dy
1−x2 y 2
=   a
√ dx
1−x2
With    a change of coordinates
β    − sin(θ)
(x = cos(θ)), the last integral becomes                 π − α − β.
π−α sin(θ)
=
Note that in this case, γ = 0. We are now left with the case where no vertex lies
on ∂H. To handle this, we simply reduce this to two instances of our previous
case by extending one of the line segments to a ray. We form two new triangles
and take their diﬀerence to be the area of the original triangle.

5      Various Trigonometries
We will end this lecture with a few comments on trigonometries in various
diﬀerent geometries. Those are the spherical geometry, the hyperbolic geometry
and the Euclidean geometry. In all three cases, we deal with triangles with sides
measuring a, b, and c, and having opposite angles α, β, and γ respectively.

5.1      Triangles on the Sphere
For a triangle on a sphere with radius r, the three trigonometric identities take
the following form.
sin(a/r)    sin(b/r)
Sine Rule:           =                                (1)
sin(α)      sin(β)
Cosine Rule I: cos(c/r) = cos(a/r) cos(b/r) + sin(a/r) sin(b/r) cos(γ)                     (2)
cos(α) cos(β) + cos(γ)
Cosine Rule II: cos(c/r) =                                                      (3)
sin(α) sin(β)

5.2      Triangles on the Hyperbolic Plane
To convert the previous equations into Hyperbolic form, we use the following
deﬁnition. Informally, it may be thought of as the “sphere with imaginary
r 2 (dx2 +dy 2 )
Deﬁnition Let Hr be the upper half plane with metric ds2 =                        y2        .

Note that letting r = 1, we get the usual hyperbolic plane. We now take the
three previous equations and replace r with ir. Using the identities sin(ix) =
i sinh(x) and cos(ix) = cosh(x), we obtain the following hyperbolic trigonomet-
ric identities.
sin(a/r)    sin(b/r)
Sine Rule:           =                              (4)
sin(α)      sin(β)
Cosine Rule I: cosh(c/r) = cosh(a/r) cosh(b/r) + sinh(a/r) sinh(b/r) cos(γ)
(5)
cos(α) cos(β) + cos(γ)
Cosine Rule II: cosh(c/r) =                                    (6)
sin(α) sin(β)

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5.3    Trigonometry on the Euclidean Plane
We can convert the above equations to Euclidean trigonometric equations by
taking the second order Taylor expansion of sin, cos, sinh, and cosh and letting
a, b, c, r → ∞. Equation (4) becomes

a        b
=                                     (7)
sin(α)   sin(β)

which is the standard Sine Rule. Likewise, equation (5) becomes

c2 = a2 + b2 − 2ab cos(γ)                         (8)

which is the standard Cosine Rule. As mentioned earlier, there is no analogue
to the Cosine Rule II.

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